IR

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Infrared Spectroscopy Part 1

Lecture Supplement:

Take one handout from the stage

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Midterm Exam 1

•Date: Monday April 30•Time: 5:00-6:50 PM•Topics: Molecular Structure: Introduction and Review Carbohydrates•Location: last name A-H in Franz 1178 last name I-Sh in Haines 39 last name Si-Z in PAB 1425

Question and Answer Session•Lecture time, Monday April 30•Submit questions to [email protected]•Label as “Question for Q&A”•Deadline for possible inclusion: noon Sunday April 29

•Extra Office Hours (Steve J): Sunday 2-4 PM, Young Hall 3077F

mailto:[email protected]

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Infrared Spectroscopy (IR)Molecular Vibrations

Fundamental principleAbsorption of photons causes changes in molecular vibrations

Molecular Vibrations•Bonded atoms move around in space•Very fast: one vibration cycle ~10-15 seconds

Stretching (H-Cl)•Atoms move along bond axis

Bending (H-O-H)•Motion not along bond axis•Less important than stretching

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Molecular Vibrations

Vibration energy• vibration energy average bond length

Ground statelower energy

add energy

Excited statehigher energy

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Molecular VibrationsVibration energy• Vibrational energy is quantized (only certain energy values are possible)

Excited vibrational state

Ground vibrational state

E = h

For bond vibrations:= dependent on bond = ~5 kcal mol-1

= lower energy than red light photons = infrared photons

= stretching frequency

Vib

ratio

nal s

tate

ene

rgy

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The Infrared Spectrum

Many photons absorbed

Spectrum = plot of photon energy versus photon quantity

Num

ber

of p

hoto

ns a

bsor

bed

Stretching frequencyProportional to photon energy

Typical infrared spectrum:

Fewphotons

absorbed

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Molecular Structure from IR SpectrumHow does spectrum give information about molecular structure?•Structure controls number of photons absorbed•Structure controls stretching frequency

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Structure versus Photon Quantity

Chance of photon absorption controlled by change in dipole moment ()

Intensity of IR peak Vector sum of bond dipoles

+ X Y -

Useful approximationConsider only one bond

From quantum mechanics:

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Absorption Intensity versus Bond Dipoles

•Bond dipole ~ (magnitude of electronegativity difference) x (bond length)•EN dipole• bond length dipole• bond dipole absorption

In practical terms:•Highly polar bond strong peak•Symmetrical (nonpolar) or nearly symmetrical bond peak weak or absent

+ X Y -

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Absorption Intensity versus Bond Dipoles

C=O peak strongH3C CH3

O

CH3H3C

H3C CH3

C=C peak absent (or maybe weak)

HH

H3C CH3

C=C peak present but weak

Examples:

Caution!•Weak peaks not always discernable•Be careful when excluding symmetrical functional groups base on absence of peak

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Structure versus Stretching FrequencyHooke’s Law (1660)•Stretching frequency of two masses on a spring

atoms bond

Stretching frequency =1

2cf

mA + mB

mAmB

1/2

bond order

stretching frequencyincreasing

spring stiffness

C-CC=CCC

atom masses

Functional groups determine IR stretching frequencies

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Characteristic Stretching FrequenciesThe Five Zones

IR spectrum divided into five zones (groups) of important absorptions

1 2 3 4 5 Fingerprint region

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Bond Stretching Frequency Shape and Intensity

Zone 1: 3700-3200 cm-1

Alcohol O-H 3650-3200 cm-1 usually strong and broad

Alkyne C-H 3340-3250 cm-1 usually strong and sharp

Amine or amide N-H 3500-3300 cm-1 medium; often broad

Zone 2: 3200-2700 cm-1

Aryl* or vinyl** sp2 C-H 3100-3000 cm-1 variable

Alkyl sp3 C-H 2960-2850 cm-1 variable

Aldehyde C-H ~2900, ~2700 cm-1 medium; two peaks

Carboxylic acid O-H 3000-2500 cm-1 usually strong; very broad

* attached to benzene ring **attached to alkene

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Zone 3: 2300-2000 cm-1

Alkyne CC 2260-2000 cm-1 sharp and variable

Nitrile CN 2260-2220 cm-1 sharp and variable

Zone 4: 1850-1650 cm-1

Ketone C=O 1750-1705 cm-1 strong

Ester C=O 1750-1735 cm-1 strong

Aldehyde C=O 1740-1720 cm-1 strong

Carboxylic acid C=O 1725-1700 cm-1 strong

Amide C=O 1690-1650 cm-1 strong

C=O frequencies 20-40 cm-1 lower when conjugated to a pi bond

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Zone 5: 1680-1450 cm-1

Alkene C=C 1680-1620 cm-1 variable

Benzene C=C~1600 and

1500-1450 cm-1

variable;

1600 cm-1 often two peaks

•Complete table: Thinkbook, inside front cover

What do I have to know?•Functional groups in each zone Learn by working lots of problems•Do not memorize stretching frequencies; table given on exam

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Guided Tour of Functional GroupsTerminal Alkyne

C CH CH2CH2CH2CH3

C-C

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Guided Tour of Functional GroupsTerminal Alkene

CC

H

CH2CH2CH2CH3

H

H

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Guided Tour of Functional GroupsAlcohol

H O CH2CH2CH2CH2CH2CH3

broad

C-O

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Guided Tour of Functional GroupsKetone

CH3C CH2CH3

O

very

str

ong

1718 cm-1

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Infrared Spectroscopy Part 2

Lecture Supplement:

Take one handout from the stage

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Infrared Spectroscopy Part 1 Summary

•Infrared photons cause excitation of molecular vibrations•Photon absorption probability higher with more polar bonds•Energy of photons absorbed depends on:

} Functional groups

•IR spectrum divided into five zones•Each zone analyzed for absence or presence of functional groups•Stretching frequency, peak shape both important

Bond orderMasses of atoms bonded

Alcohol O-H usually gives broad peakC=O stretch gives strong peak

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Guided Tour of Functional GroupsKetone (again)

CH3C CH2CH3

O

very

str

ong

1718 cm-1

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Guided Tour of Functional GroupsAldehyde

CH CH2CH2CH2CH2CH3

O

~2900 cm-1

usually obscured

very

str

ong

1718 cm-1

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Guided Tour of Functional GroupsKetone with Alkene Conjugation

CH3C C

C

O

H

H

HConjugation with pi bond lowers

C=O stretch by 20-40 cm-1

1720 cm-1

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Guided Tour of Functional GroupsEster

CCH3O CH2CH2CH2CH2CH3

O

1743 cm-1

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Guided Tour of Functional GroupsCarboxylic Acid

CO CH2CH2CH2CH2CH3

O

H

very broad

1711 cm-1

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Guided Tour of Functional GroupsBenzene Ring

H3C

HH

H H

H

May be two peaks

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Five Zone IR Spectrum AnalysisExample #1: C6H12O2

1700 cm-1

Step 1: Calculate DBEDBE = C - (H/2) + (N/2) + 1 = 6 - (12/2) + (0/2) +1 = 1 One ring or one pi bond

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1700 cm-1

Step 2: Analyze IR Spectrum•Zone 1 (3700-3200 cm-1)

PresentAbsent - no N in formulaAbsent - not enough DBE

Alcohol O-H:N-H:

C-H:

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1700 cm-1

•Zone 2 (3200-2700 cm-1)

Aryl/vinyl sp2 C-H:Alkyl sp3 C-H:Aldehyde C-H:

Carboxylic acid O-H:

Probably not (not enough DBE)

Absent - no 2700 cm-1

Absent - not broad enough

Present

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1700 cm-1

•Zone 3 (2300-2000 cm-1)

Alkyne CC:Nitrile CN:

Absent - no peaks; not enough DBEAbsent - no peaks; not enough DBE

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1700 cm-1

•Zone 4 (1850-1650 cm-1)

C=O:Possibilities: ketone ester - not enough oxygens aldehyde - no 2700 cm-1 peak carboxylic acid - zone 2 not broad amide - no nitrogen

Present @ 1700 cm-1

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1700 cm-1

•Zone 5 (1680-1450 cm-1)Benzene ring:Alkene C=C:

Absent - no peak ~1600 cm-1; not enough DBEAbsent - no peak ~1600 cm-1; not enough DBE

Actual structure:OHO

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Five Zone IR Spectrum AnalysisExample #2: C8H7N

Step 1: Calculate DBEDBE = C - (H/2) + (N/2) + 1 = 8 - (7/2) + (1/2) +1 = 6 Six rings and/or pi bonds Possible benzene ring

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Step 2: Analyze IR Spectrum•Zone 1 (3700-3200 cm-1)

Absent - no oxygen in formulaAbsent - peaks too smallAbsent - peaks too small

Alcohol O-H:N-H:

C-H:

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•Zone 2 (3200-2700 cm-1)

Aryl/vinyl sp2 C-H:Alkyl sp3 C-H:Aldehyde C-H:

Carboxylic acid O-H:

Present - peaks > 3000 cm-1

Present - peaks < 3000 cm-1

Absent - no 2700 cm-1; no C=O in zone 4Absent - not broad enough; C=O in zone 4

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•Zone 3 (2300-2000 cm-1)

Alkyne CC:Nitrile CN:

PossiblePossible }

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•Zone 4 (1850-1650 cm-1)

C=O: Absent - no peak; no oxygen in formula

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•Zone 5 (1680-1450 cm-1)Benzene ring:Alkene C=C:

Present - peaks ~1600 cm-1 and ~1500 cm-1

triple bondAbsent - not enough DBE for alkene plus benzene plus

Actual structure: CH2CC N

IR

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